This problem is part of one of the free response questions on the 2011 AP Physics C Mechanics exam [PDF of free-response questions] that was administered last week.

(Sorry the problem statement is so long; I wanted to make sure I'm not inadvertently changing the meaning by shortening it.)

A projectile with mass $m$ and velocity $v_x$ is fired horizontally into a block of wood that is clamped to a tabletop so that it cannot move. The projectile travels a distance $d$ into the block before it stops.

(d) Derive an expression for the average force $F_b$ exerted on the projectile as it comes to rest in the block.

Now a new projectile and block are used, identical to the first ones, but the block is not clamped to the table. The projectile is again fired into the block of wood and travels a new distance $d_n$ into the block while the block slides across the table a short distance $D$. Assume the following: the projectile enters the block with speed $v_x$, the average force $F_b$ between the projectile and the block has the same value as determined earlier, the average force of friction between the table and the block is $f_T$, and the collision is instantaneous so the frictional force is negligible during the collision.

(e) Derive an expression for $d_n$ in terms of $d$, $D$, $f_T$, and $F_b$, as appropriate.
(f) Derive an expression for $d_n$ in terms of $d$, the mass $m$ of the projectile, and the mass $M$ of the block.

For part (d), I simply used energy to find $F_b$ using that $E = \frac12 m v_x^2 = F_b d$.

I'm pretty sure I'm right up to this point, but part (f) is quite confusing – it seems to me that there's absolutely no way that $d_n$ can be expressed only in terms of those three quantities because it needs to depend on the frictional force between the block and the table. Can anyone explain if I'm missing something? If I'm not missing anything, what is the most likely original intention of the question-writers?

Ben, In your solution to part e you seem to assume that the block has come to rest after moving the distance, D. In my interpretation of this description, the time it takes for the projectile to penetrate the block to a distance, dn is the same time it takes the block to travel a distance, D. But it doesn't indicate that at this time the block has come to rest.
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user3541May 14 '11 at 22:56

1 Answer
1

You certainly have a right to be upset about a poorly-written question. The question talks about "average force" without defining what that means. Usually, I would interpret "average force" to mean a time-average, but it looks like the person who wrote the problem intended it to mean an average over space.

An average over space is an ill-defined concept in this case. When the bullet enters the block and the block starts sliding backwards, are we averaging over the relative motion of the bullet through the block, or over the motion of the bullet relative to the table top? Again, from context I'll infer it's an average over the path of the bullet relative to the block.

That being said, you can solve the problem by considering where the energy goes. In the first case with the clamped block, all the energy goes into driving the bullet into the block.

In the second case, some of the energy goes into stopping the bullet during the collision, just as before, but some of the energy goes into kinetic energy of the block+bullet system after the collision (and is then dissipated by friction with the table).

From $KE = p^2/2m$ and conservation of momentum, we see that the fraction of the bullet's original kinetic energy that goes into the block+bullet's kinetic energy immediately after impact is $m/(M+m)$.

That's the fraction of the kinetic energy that does not go into stopping the bullet, so the bullet's penetration depth goes down by a factor of one minus that, so

$$d_n = d \left(1 - \frac{m}{M+m}\right)$$

edit: I simplified this a bit after re-reading the part about how we're supposed to consider the collision and the slowing of the block/bullet system due to friction as separate processes.

I thought the intention was reasonably clear: assume that the projectile decelerates at a constant rate until its velocity is stopped. Obviously the shape of the deceleration versus time/interaction time curve must be specified, otherwise the problem is not well specified (i.e. the unknown shape of this curve will affect the result). Constant deceleration is the simplest possible functional form, and likely makes the algebra easily tractable.
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Omega CentauriMay 15 '11 at 3:27

@Omega I guess I was pointing out that "average force" does not mean the same as "constant force". I agree that if you assume the force is constant you'll be fine.
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Mark EichenlaubMay 15 '11 at 5:44

This is a great answer, thank you very much!
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Ben AlpertMay 15 '11 at 18:18